– LA FORMULA DI TAYLOR COL RESTO DI LAGRANGE Sia Pn(x ...
– LA FORMULA DI TAYLOR COL RESTO DI LAGRANGE Sia Pn(x ...
– LA FORMULA DI TAYLOR COL RESTO DI LAGRANGE Sia Pn(x ...
You also want an ePaper? Increase the reach of your titles
YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.
2 <strong>LA</strong> FORMU<strong>LA</strong> <strong>DI</strong> <strong>TAYLOR</strong> <strong>COL</strong> <strong>RESTO</strong> <strong>DI</strong> <strong>LA</strong>GRANGE<br />
In pratica la ϕ si ottiene dalla (R) mettendo la variabile t al posto del numero<br />
fissato c, e sottraendo la quantità a (x−t)n+1<br />
(x−c) n+1.<br />
Chiaramente ϕ è continua in ]α, β[ e dotata in ogni punto t ∈]α, β[ distinto da c<br />
di derivata prima ϕ ′ (t). Osserviamo ora che si ha ϕ(x) = ϕ(c) = 0. Infatti<br />
ϕ(x) = f(x) −<br />
e<br />
n�<br />
k=0<br />
f (k) (x)<br />
(x − x)<br />
k!<br />
k (x − x)n+1<br />
− a =<br />
(x − c) n+1<br />
= f(x) − f(x) − f ′ (x)(x − x) − f ′′ (x)(x − x) 2 − · · · − f (n) (x)(x − x) n = 0,<br />
ϕ(c) = f(x) −<br />
= f(x) −<br />
n�<br />
k=0<br />
n�<br />
k=0<br />
f (k) (c)<br />
(x − c)<br />
k!<br />
k (x − c)n+1<br />
− a =<br />
(x − c) n+1<br />
f (k) (c)<br />
(x − c)<br />
k!<br />
k − a = 0<br />
per la (R). Per il teorema di Rolle esiste così un punto ξ interno all’intervallo<br />
di estremi c ed x tale che ϕ ′ (ξ) = 0. Calcoliamo, per t ∈]α, β[, t �= c, ϕ ′ (t). Si ha<br />
� ′<br />
ϕ ′ (t) =<br />
�<br />
f(x) −<br />
n�<br />
k=0<br />
f (k) (t)<br />
(x − t)<br />
k!<br />
k (x − t)n+1<br />
− a<br />
(x − c) n+1<br />
�<br />
= 0 − f(t) + f ′ (t)(x − t) + f ′′ (t)<br />
2 (x − t)2 + f ′′′ (t)<br />
(x − t)<br />
3!<br />
3 + · · · + f(n) (t)<br />
(x − t)<br />
n!<br />
n<br />
� ′<br />
+<br />
�<br />
(x − t)n+1<br />
− a<br />
(x − c) n+1<br />
� ′<br />
=<br />
= − f ′ �<br />
(t) − f<br />
����<br />
•<br />
′′ (t)(x − t) − f<br />
� �� �<br />
••<br />
′ � � ′′′ f (t)<br />
(t) − (x − t)<br />
���� 2<br />
•<br />
2<br />
−<br />
� �� �<br />
•••<br />
f ′′ � � ′′′′ (t) f (t)<br />
2 (x − t) − (x − t)<br />
�<br />
2<br />
�� �<br />
3!<br />
••<br />
3<br />
+<br />
� �� �<br />
••••<br />
− f ′′′ (t)<br />
3 (x − t)<br />
3!<br />
2<br />
� � (n+1) f (t)<br />
− · · · − (x − t)<br />
� �� �<br />
n!<br />
•••<br />
n − f(n) (t)<br />
n (x − t)<br />
n!<br />
n−1<br />
�<br />
(x − t)<br />
+ a (n + 1)<br />
� �� �<br />
•<br />
�<br />
•<br />
��<br />
· · · •<br />
�<br />
n<strong>–</strong>volte<br />
n<br />
=<br />
(x − c) n+1<br />
= − f(n+1) (t)<br />
(x − t)<br />
n!<br />
n (x − t)<br />
+ a (n + 1)<br />
n<br />
=<br />
(x − c) n+1<br />
= −<br />
Cosìda<br />
(n + 1)(x − t)n<br />
(x − c) n+1<br />
� �<br />
(n+1) n+1<br />
f (t)(x − c)<br />
− a .<br />
(n + 1)!<br />
ϕ ′ (ξ) = −<br />
segue necessariamente<br />
(n + 1)(x − ξ)n<br />
(x − c) n+1<br />
� �<br />
(n+1) n+1<br />
f (ξ)(x − c)<br />
− a = 0<br />
(n + 1)!<br />
a = f(n+1) (ξ)(x − c) n+1<br />
(n + 1)!<br />
=